What are the solutions to the equation -x^2+6x+18=0 ?

The solutions to the equation -x^2+6x+18=0 are x=-3(sqrt(3)-1),x=3(1+sqrt(3))

Find the zeros of -x^2+6x+18=0

The zeros of -x^2+6x+18=0 are x=-3(sqrt(3)-1),x=3(1+sqrt(3))

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- x^{2} + 6 x + 18 = 0

x = - 3 (3 - 1), x = 3 (1 + 3)

Decimal Notation

x = - 2.19615 \dots, x = 8.19615 \dots

Solution steps

- x^{2} + 6 x + 18 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 ( - 1 ) \cdot 18}{2 ( - 1 )}

6^{2} - 4 (- 1) \cdot 18 = 63

x_{1, 2} = \frac{- 6 \pm 6 3}{2 ( - 1 )}

Separate the solutions

x_{1} = \frac{- 6 + 6 3}{2 ( - 1 )}, x_{2} = \frac{- 6 - 6 3}{2 ( - 1 )}

x = \frac{- 6 + 6 3}{2 ( - 1 )} : - 3 (3 - 1)

x = \frac{- 6 - 6 3}{2 ( - 1 )} : 3 (1 + 3)

The solutions to the quadratic equation are:

x = - 3 (3 - 1), x = 3 (1 + 3)

t, d = \frac{1}{2} g t^{2}

- 5 x^{2} - 80 x - 140 = 0

6 x^{2} - 4 x = 2

6 m^{2} + 26 m = 20

3 b^{2} = 1

What are the solutions to the equation -x^2+6x+18=0 ?
The solutions to the equation -x^2+6x+18=0 are x=-3(sqrt(3)-1),x=3(1+sqrt(3))
Find the zeros of -x^2+6x+18=0
The zeros of -x^2+6x+18=0 are x=-3(sqrt(3)-1),x=3(1+sqrt(3))